Question 120363
Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{2*x^2+7*x+1=0}}} ( notice {{{a=2}}}, {{{b=7}}}, and {{{c=1}}})





{{{x = (-7 +- sqrt( (7)^2-4*2*1 ))/(2*2)}}} Plug in a=2, b=7, and c=1




{{{x = (-7 +- sqrt( 49-4*2*1 ))/(2*2)}}} Square 7 to get 49  




{{{x = (-7 +- sqrt( 49+-8 ))/(2*2)}}} Multiply {{{-4*1*2}}} to get {{{-8}}}




{{{x = (-7 +- sqrt( 41 ))/(2*2)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-7 +- sqrt(41))/(2*2)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (-7 +- sqrt(41))/4}}} Multiply 2 and 2 to get 4


So now the expression breaks down into two parts


{{{x = (-7 + sqrt(41))/4}}} or {{{x = (-7 - sqrt(41))/4}}}



Now break up the fraction



{{{x=-7/4+sqrt(41)/4}}} or {{{x=-7/4-sqrt(41)/4}}}






So these expressions approximate to


{{{x=-0.149218940641788}}} or {{{x=-3.35078105935821}}}



So our solutions are:

{{{x=-0.149218940641788}}} or {{{x=-3.35078105935821}}}


Notice when we graph {{{2*x^2+7*x+1}}}, we get:


{{{ graph( 500, 500, -13.3507810593582, 9.85078105935821, -13.3507810593582, 9.85078105935821,2*x^2+7*x+1) }}}


when we use the root finder feature on a calculator, we find that {{{x=-0.149218940641788}}} and {{{x=-3.35078105935821}}}.So this verifies our answer